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Question

A point moves in xy-plane according to equation x=at,y=at( 1-bt)where a and b are positive constants and t is time. The instant at which velocity vector is at π/4 with acceleration vector is given by

A
1/a
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B
1/b
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C
1/a+1/b
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D
(a+b)/(a2+b2)
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Solution

The correct option is A 1/bx=aty=at(1−bt)We have to find the time when the angle between acceleration vector and velocity vector will be π/4Now,vx=dxdt=avy=dydt=a(1−2bt)So, the velocity vector will be →v=a^i+a(1−2bt)^j the acceleration vector will be →A=d→vdt=−2ab^jSo, as you can see the picture acceleration →A is in −Y direction and velocity →v makes an angle π/4So,tan(π/4)=a|a(1−2bt)|or, |1−2bt|=1or, 1−2bt=±1or,t=0 or1bBut, when t=0, the Y-component of velocity will be in positive direction. So the condition will not work.Hence, our answer is t=1bThe correct answer is Option:B

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